physics on the road - lesson 7
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Speeding up, slowing down?
Speeding up, slowing down
Speeding up, slowing down
Physics on the RoadLesson 7
LI…Describe uniform acceleration.Use kinematic equations to
make predictionsInterpret motion graphs
Constant accelerationWhen something gains speed at a constant
rate this is called uniform acceleration. Lets begin with motion in a straight line
vv∆vv+∆v∆v
∆t∆t∆t t=0
vv +∆v
v +2∆v
Traffic Incident
A policeman using a video camera and radar claims that a car starting from some lights reached 90 km/h when it passed a shop 150m away 12 s later. The driver claims this is impossible.
Distance =150 mInitial velocity = 0 m/sFinal velocity = 90 km/h
= 25 m/sTime = 12 s
Check 1: many cars advertise a 0-100 km/h of less than 10s
Check 2: The car would need to gain 25/12=2.08m/s each of the 12 seconds or 2.08 m/s/s
metres per second per second is the unit of acceleration and is abbreviated to ms-2.
Traffic Incident cont…A policeman using a video camera and radar claims that a car starting from some lights reached 90 km/h when it passed a shop 150m away 12 s later. The driver claims this is impossible.
Distance =150 mInitial velocity = 0 m/sFinal velocity = 25 m/sTime = 12 s
Car accelerating to 25m/s in 12s
0
5
10
15
20
25
0 1 2 3 4 5 6 7 8 9 10 11 12
Time /s
speed /
m/s
∆v
∆t
acceleration = change of velocity time
acceleration = 25 = 2.08 ms-2
12
avg. speed = change of distance time
avg. speed = 150 = 12.5 ms-1
12
avg. speed = 0 + 25 = 12.5 ms-1
2
distance = avg. speed x time = 12.5 x 12= 150m
Linking to distance time graphs Draw this velocity
time graph Use your graph and
the relationship between average speed and distance travel to complete a distance time table.
Plot the distance time graph
Car accelerating to 25m/s in 12s
0
5
10
15
20
25
0 1 2 3 4 5 6 7 8 9 10 11 12
Time /s
speed /
m/s
distance = avg. speed x time = 12.5 x 12= 150m
Time /s Distance /m
12 150
Distance travelled is the areaunder a velocity time graph
Kinematic equations
acceleration = change of velocity time
acceleration = final velocity – initial velocity time
a = v – ut
at= v – u
v= u + at
Car accelerating to 25m/s in 12s
0
5
10
15
20
25
0 1 2 3 4 5 6 7 8 9 10 11 12
Time /s
speed /
m/s
Distance travelled is the area under the graph
Area of triangle = ½ x base x height = ½ x time x change in velocity
= ½ x 12 x 25 = 150 m
First equationWhat if the object wasn’t originally stationary?
More kinematic equations
velocity time graph
012345678
0 1 2 3 4
time (s)
velo
city
(m
/s)
The distance travelled is the area under the graph
Area A + Area B
A
B
u
v
u x t + ½ x t x (v-u)
but… at = v – uso…
s = ut + ½ at2
2nd
equation
Third equation of motion
v= u + att= v – u a
average velocity= u + v 2
Distance = average velocity x timeSo….
s=u+v t 2s = (u+v )x (v-u) 2 as = v2 – u2
2a
(u+v)(v-u)= v2 – u2
v2 = u2 + 2as3rd equation
1st equation
Newton’s equations of motion
v= u + at v2 = u2 + 2ass = ut + ½ at2
Use to computefinal velocity from acceleration and time.
Use to computedistance from acceleration and time.
Use to computeFinal velocity from acceleration and distance.
v – final velocity (ms-1) t – time (s)u – initial velocity (ms-1) s –distance/displacement (m) a – acceleration (ms-2)
Questions
v= u + at
v2 = u2 + 2as
s = ut + ½ at2
A car travelling at 10 ms-1 accelerates at 3ms-2 for 5 seconds what is the car’s final velocity?.
How far will the car travel whilst accelerating?
A coin is dropped from the Centre Point tower in Sydney. The tower is 309m high. Acceleration due to gravity is 10 ms-2. What speed does the coin hit the ground? What assumption have you made?
Q 1 a Q 1 b
Q 2